Department of Electrical and Computer Engineering The University of Texas at Austin

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1 Department of Electrical and Computer Engineering The University of Texas at Austin EE 306, Fall 2011 Yale Patt, Instructor Faruk Guvenilir, Milad Hashemi, Jennifer Davis, Garrett Galow, Ben Lin, Taylor Morrow, Stephen Pruett, and Jee Ho yoo, TAs Exam 2, November 2, 2011 Name: Problem 1 (15 points): Problem 2 (10 points): Problem 3 (20 points): Problem 4 (15 points): Problem 5 (15 points): Problem 6 (25 points): Total (100 points): Note: Please be sure that your answers to all questions (and all supporting work that is required) are contained in the space provided. Note: Please be sure your name is recorded on each sheet of the exam. I will not cheat on this exam. Signature GOOD LUCK!

2 Name: Problem 1. (15 points): Part a. (5 points): An LC-3 computer, during execution of a program, encounters during a period of 19 clock cylces the following states of the state machine in sequence: 18, 33, 35, 32, 2, 25, 27, 18, 33, 35, 32, 6, 25, 27, 18, 33, 35, 32, 1 List the opcodes of the instructions executed in the order executed. Use as many entries as you need: Part b. (5 points): Construct the symbol table for the block of code on the left:.oig x4000 ADD 1,2,3 DIVIDE AND 1,2,3 JMP HEE HELP1.BLKW 4 HELP2.STINGZ "HELP" HEE.FILL xf025.end Symbol Table: Symbol Address 2

3 Name: Part c. (5 points): The LC-3 state machine shows that after Decode (state 32), the TAP instruction requires three clock cycles to finish execution, represented as states 15, 28, and 30. These three states are reproduced below (on the left). Note that 7 gets loaded with PC many times, depending on how many cycles it takes to read memory. A suggestion: Move 7 < PC from state 28 to state 30 so it will get done once (as shown on the right). Before After TAP TAP MA < ZEXT[I[7:0]] MA < ZEXT[I[7:0]] MD < M[MA] 7 < PC 28 MD < M[MA] PC < MD PC < MD 7 < PC to 18 to 18 Will the suggested new implemenation work on our existing data path? EXPLAIN in 15 words or fewer. 3

4 Name: Problem 2. (10 points): We wish to add a new TAP service routine, which will be called by the instruction TAP x9a. The new trap routine will wait for someone to type a lower case letter, then echo on the screen the corresponding capital letter. Assume the user will not type anything except a lower case letter. The assembly language code for this trap service routine is shown below:.oig x2055 ST 1, Save1 ST 0, Save0 TAP x20 LD 1, A TAP x21 LD 1, Save1 LD 0, Save0 JMP 7 Save1.BLKW 1 Save0.BLKW 1 A.FILL.BLKW 1.END Part a: In order for TAP x9a to call this service routine, what memory location must contain what value. Address: Value: Part b: Fill in the missing information in the assembly language program. i.e, the three missing instructions, the one missing label, and the operand of the.fill pseudo-op. 4

5 Name: Problem 3. (20 points): The following LC-3 assembly language program:.oig x3000 AND 2, 2, #0 AND 6, 6, #0 ADD 2, 2, #1 TOP ADD 3, 2, #0 ADD 4, 1, #0 SEACH ADD 3, 3, 3 ADD 4, 4, #-1 Bp SEACH AND 5, 3, 0 Bz NEXT ADD 6, 6, 2 NEXT ADD 2, 2, 2 END Bzp TOP ST 6, ESULT HALT ESULT.BLKW 1.END What does it do (in twenty words or fewer)? Please be BIEF but PECISE. You can assume that some of the registers will already contain numbers that are relevant to the program. What is the function of 0? For what range of input values does the program function as you ve described above? What is the function of 1? For what range of input values does the program function as you ve described above? What is the function of 6? For what range of input values does the program function as you ve described above? 5

6 Name: Problem 4. (15 points): The following LC-3 assembly language program executes to completion..oig x3000 AND 0, 0, #0 ADD 1, 0, #1 ADD 3, 0, #10 LEA 6, ESULTS LOOP ADD 2, 0, 1 ADD 0, 1, #0 ADD 1, 2, #0 STOE ST 1, 6, #0 LD 2, STOE ADD 2, 2, #1 ST 2, STOE ADD 3, 3, #-1 Bp LOOP HALT ESULTS.BLKW 10.END Part a. (9 points): After the program has halted, what values are contained (in decimal) in the ten consecutive memory locations starting at the memory location labeled ESULTS? Part b. (6 points): The three instructions inside the box above can be replaced by a single instruction while preserving the functionality of the program. What is that single instruction? 6

7 Name: Problem 5. (15 points): Let s use the unused opcode 1101 to specify a new instruction. We will require 3 states after decode (state 32) to complete the job. The control signals required to carry out the work of the new instruction are shown below. All control signals not shown below are 0. BEN < I[11] & N + I[10] & Z + I[9] & P [I[15:12]] LD.MD = 1 GateALU = 1 S1MUX = I[8:6] ALUK = ADD State 13 LD.MA = 1 GateMAMUX = 1 S1MUX = I[11:9] ADD1MUX = Base ADD2MUX = #0 MAMUX = ADDE OUTPUT State A In 20 words or fewer, what does the new instruction do? MIO.EN = 1.W = WITE To State 18 State B The instruction may have one or two potential formats. Fill out one or both as you deem appropriate

8 Name: Problem 6. (25 points): An LC-3 program is executing on the LC-3 Simulator when a breakpoint is encountered, and the Simulator stops. At that point, the contents of several registers are as shown in the first row of the table. After the run button is subsequently pushed, the next four instructions that are executed, none of which are an STI or LDI, produce the values shown in the table, two rows of the table per instruction executed. The first row of each pair shows the contents after the fetch phase of the corresponding instruction, and the second row of each pair after that instruction completes. Note that some values are missing, and are presented by letters A, B, C, D, E, F, G, H, I, and J. PC MA MD I x1800 x7fff x2211 xbffe x31ff x2233 x5177 x3211 x21ff x5233 x3177 x2211 A x1800 B B x31ff x2233 x5177 x3211 x21ff x5233 x3177 x2211 A x1800 B B x31ff x2233 x5177 x3211 x21ff C x3177 x2211 D A E E x31ff x2233 x5177 x3211 x21ff C x3177 x2211 D F G E x31ff x2233 x5177 x3211 x21ff C x3177 x2211 H D I I x31ff x2233 x5177 x3211 x21ff C x3177 x2211 F D I I x31ff x2233 x5177 x3211 x21ff C x3177 x2211 A F J J x31ff x2233 x5177 x3211 x21ff C x3177 x2211 A F J J x31ff x2233 x5177 x3211 x223a C x3177 x2211 Your job: Determine the values of A, B, C, D, E, F, G, H, I, and J. Note that some of the values may be identical. A B C D E F G H I J x F 8

9 18 MA < PC PC < PC + 1 [INT] MD < M To 49 (See figure C.7) I < MD To 8 (See figure C.7) TI ADD BEN< I[11] & N + I[10] & Z + I[9] & P [I[15:12]] 1101 B To 13 D< S1+OP2* set CC D< S1&OP2* set CC PC< MD 1 5 D< NOT(S) 9 set CC 15 MA< ZEXT[I[7:0]] 28 MD< M[MA] 7< PC D< PC+off9 set CC AND NOT 6 TAP MA< PC+off9 LEA LD LD LDI STI ST ST MA< B+off6 10 MA< PC+off9 MD< M[MA] MA< MD MA< PC+off9 MD< M[MA] MA< MD 29 MA< B+off6 31 JS JMP 7 MA< PC+off9 3 0 [BEN] 1 PC< PC+off PC< Base [I[11]] 1 0 7< PC PC< PC+off < PC PC< Base MD< M[MA] D< MD set CC 27 MD< S M[MA]< MD 23 NOTES B+off6 : Base + SEXT[offset6] PC+off9 : PC + SEXT{offset9] PC+off11 : PC + SEXT[offset11] *OP2 may be S2 or SEXT[imm5] 9

10 MEMOY OUTPUT INPUT DS DD KBD ADD. CTL. LOGIC GateMD MD LD.MD INMUX MA LD.MA 2 KBS MIO.EN MEM.EN.W MIO.EN SEXT SEXT SEXT SEXT [5:0] [8:0] [10:0] +1 GateMAMUX ALU B A GateALU S2MUX PC + I ZEXT N Z P LOGIC S2 OUT S1 OUT EG FILE [7:0] 2 PCMUX LD.CC GatePC LD.PC LD.I MAMUX ALUK [4:0] 0 ADD1MUX 2 ADD2MUX S1 S2 LD.EG D 3 CONTOL 10

11 I[11:9] I[11:9] 110 D I[8:6] S DMUX S1MUX (a) (b) I[11:9] N Z P Logic BEN (c) 11

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